Q.The integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}$ is:
(A) $\frac{x}{e^x}$
(B) $\frac{e^x}{x}$
(C) $xe^x$
(D) $e^x$

Showing the 12 most recent of 33 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The general solution of the equation $\dfrac{dy}{dx}+\dfrac{1}{x}y=\dfrac{1}{x}e^x$ is (A) $y=xe^x+c$ (B) $y=xe^x+ce^{-x}$ (C) $y=\dfrac{e^x+c}{x}$ (D) $y=\dfrac{e^{-x}+cx}{x}$

   ›Reveal solutionSolution

   This is a standard linear first-order ODE solved via an integrating factor; the answer is (C).

   Concept and Intuition

   An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu=e^{\int P\,dx}$, which makes the left side a perfect derivative $\dfrac{d}{dx}(\mu y)$.

   Step-by-Step Solution

   1. Here $P(x)=\dfrac1x$, $Q(x)=\dfrac1x e^x$.
   2. Integrating factor: $\mu=e^{\int \frac1x dx}=e^{\log x}=x$.
   3. Multiply the ODE by $x$: $x\dfrac{dy}{dx}+y=e^x$, i.e. $\dfrac{d}{dx}(xy)=e^x$.
   4. Integrate both sides: $xy=e^x+c$. …

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2. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.An integrating factor of the differential equation $(x^2+1)\dfrac{dy}{dx} + xy = x^3$ is (A) $\dfrac{x}{1+x^2}$ (B) $\dfrac{1}{2}\log(1+x^2)$ (C) $\sqrt{1+x^2}$ (D) $e^{\log(1+x^2)}$

   ›Reveal solutionSolution

   Writing the ODE in standard linear form $\frac{dy}{dx}+Py=Q$ and computing $e^{\int P\,dx}$ gives the integrating factor $\sqrt{1+x^2}$.

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ is solved by multiplying through by the integrating factor $\mu(x)=e^{\int P(x)\,dx}$, chosen precisely so that $\dfrac{d}{dx}\big[\mu(x)y\big]=\mu(x)Q(x)$. So the entire method hinges on correctly identifying $P(x)$ after dividing the equation into standard form.

   Step-by-Step Solution

   1. Start with $(x^2+1)\dfrac{dy}{dx}+xy=x^3$.
   2. Divide by $(x^2+1)$ to reach standard form: $\dfrac{dy}{dx}+\dfrac{x}{x^2+1}y=\dfrac{x^3}{x^2+1}$.
   3. Here $P(x)=\dfrac{x}{x^2+1}$.
   4. Compute $\int P(x)\,dx=\displaystyle\int \dfrac{x}{x^2+1}dx=\dfrac12\log(x^2+1)$ (substituting $u=x^2+1$). …

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3. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\log y$ is an integrating factor of $\dfrac{dx}{dy}+P(y)x=Q(y)$, then $P(y)=$ (A) $\dfrac{1}{y+\log y}$ (B) $\dfrac{y}{\log y}$ (C) $\dfrac{\log y}{y}$ (D) $\dfrac{1}{y\log y}$

   ›Reveal solutionSolution

   Working backward from the known integrating factor $\log y=e^{\int P(y)dy}$ and differentiating recovers $P(y)=\dfrac{1}{y\log y}$.

   Concept and Intuition

   For a linear ODE $\dfrac{dx}{dy}+P(y)x=Q(y)$ (linear in $x$, with $y$ as the independent variable), the integrating factor is $\mathrm{IF}=e^{\int P(y)\,dy}$. If we're told what the integrating factor equals, we can reverse-engineer $P(y)$ by taking logs and then differentiating.

   Step-by-Step Solution

   1. Standard fact: $\mathrm{IF}=e^{\int P(y)\,dy}$.
   2. Given $\mathrm{IF}=\log y$ (i.e. $\log y$), set $e^{\int P(y)dy}=\log y$.
   3. Take the natural log of both sides: $\displaystyle\int P(y)\,dy=\log(\log y)$.
   4. Differentiate both sides with respect to $y$ (by the Fundamental Theorem of Calculus, the left side gives back $P(y)$): $P(y)=\dfrac{d}{dy}\left[\log(\log y)\right]$.
   5. By the chain rule: $\dfrac{d}{dy}\log(\log y)=\dfrac{1}{\log y}\cdot\dfrac{d}{dy}(\log y)=\dfrac{1}{\log y}\cdot\dfrac1y=\dfrac{1}{y\log y}$. …

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4. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $x\,dy+(y+y^2x)\,dx=0$ and $y=1$ at $x=1$, then (A) $y=\dfrac{x}{1+\log x}$ (B) $y=\dfrac{1+\log x}{x}$ (C) $y=x(1+\log x)$ (D) $y=\dfrac{1}{x(1+\log x)}$

   ›Reveal solutionSolution

   A Bernoulli equation in disguise; the substitution $v=1/y$ linearizes it, and the initial condition $y(1)=1$ pins down the constant to give $y=\dfrac{1}{x(1+\log x)}$.

   Concept and Intuition

   An equation of the form $\dfrac{dy}{dx}+P(x)y=Q(x)y^n$ is a Bernoulli equation. Dividing by $y^n$ and substituting $v=y^{1-n}$ converts it into a linear first-order ODE in $v$, which can then be solved with the standard integrating-factor method.

   Step-by-Step Solution

   1. Given $x\,dy+(y+y^2x)\,dx=0$. Divide by $dx$: $x\dfrac{dy}{dx}+y+y^2x=0$.
   2. Divide by $x$: $\dfrac{dy}{dx}+\dfrac{y}{x}=-y^2$ — a Bernoulli equation with $n=2$.
   3. Divide throughout by $y^2$: $y^{-2}\dfrac{dy}{dx}+\dfrac{1}{xy}=-1$.
   4. Let $v=y^{-1}$, so $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$, i.e. $y^{-2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}$.
   5. Substitute: $-\dfrac{dv}{dx}+\dfrac{v}{x}=-1 \Rightarrow \dfrac{dv}{dx}-\dfrac{v}{x}=1$, a linear ODE in $v$.
   6. Integrating factor $=e^{\int -1/x\,dx}=e^{-\log x}=\dfrac{1}{x}$.
   7. $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=\dfrac{1}{x} \Rightarrow \dfrac{v}{x}=\log x+C \Rightarrow v=x(\log x+C)$. …

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5. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} + P(x)y = Q(x)$ is a solution of the differential equation (A) $\dfrac{dy}{dx} - P(x)y = 0$ (B) $\dfrac{dy}{dx} + P(x)y = 0$ (C) $\dfrac{dy}{dx} - \dfrac{y}{x} = P(x)$ (D) $\dfrac{dy}{dx} + \dfrac{x}{y} = P(x)$

   ›Reveal solutionSolution

   The integrating factor $\mu=e^{\int P\,dx}$ is itself a solution of the homogeneous equation $y'=P(x)y$.

   Concept and Intuition

   The whole point of the integrating factor is that $\dfrac{d}{dx}\big(\mu y\big)=\mu\dfrac{dy}{dx}+\mu P(x)y$ should equal $\mu\left(\dfrac{dy}{dx}+P(x)y\right)$, which requires $\dfrac{d\mu}{dx}=\mu P(x)$ — i.e. $\mu$ itself solves the first-order linear homogeneous ODE $y'-P(x)y=0$.

   Step-by-Step Solution

   1. The integrating factor is defined as $\mu(x)=e^{\int P(x)\,dx}$.
   2. Differentiate: $\mu'(x) = P(x)\,e^{\int P\,dx} = P(x)\mu(x)$.
   3. Rearranged: $\mu'(x) - P(x)\mu(x)=0$. …

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6. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Integrating factor of the differential equation $\sin x \dfrac{dy}{dx} - y\cos x = 1$ is (A) $\sin x$ (B) $\cos x$ (C) $\sec x$ (D) $\mathrm{cosec}\, x$

   ›Reveal solutionSolution

   Rewriting the equation in standard linear form and computing $e^{\int P\,dx}$ gives the integrating factor $\csc x$ — (D).

   Concept and Intuition

   A first-order linear ODE $\dfrac{dy}{dx}+P(x)y=Q(x)$ has integrating factor $e^{\int P(x)dx}$. The given equation must first be divided through so the coefficient of $\dfrac{dy}{dx}$ is exactly $1$.

   Step-by-Step Solution

   1. Given: $\sin x\dfrac{dy}{dx}-y\cos x=1$.
   2. Divide by $\sin x$: $\dfrac{dy}{dx}-\dfrac{\cos x}{\sin x}y=\dfrac1{\sin x}$, i.e. $\dfrac{dy}{dx}-(\cot x)y=\csc x$.
   3. Here $P(x)=-\cot x$.
   4. Integrating factor $=e^{\int -\cot x\,dx}=e^{-\log|\sin x|}=(\sin x)^{-1}=\csc x$.

   Common Mistakes …

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7. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.By multiplying with $e^{\int P\,dx}$ on both sides of the equation $\dfrac{dy}{dx} + P(x)y = Q(x)$, the left side of the equation takes the form $\dfrac{d}{dx}(y\,f(x))$, then $f(x) =$ (A) $\displaystyle\int y\,e^{\int P\,dx}\,dx$ (B) $y\,P(x)$ (C) $e^{\int P\,dx}$ (D) $P(x)\,e^{\int P\,dx}$

   ›Reveal solutionSolution

   This is the definition-check behind the standard integrating-factor method for first-order linear ODEs: the multiplier that turns the left side into an exact derivative $\frac{d}{dx}(y\cdot(\text{something}))$ is, by construction, the integrating factor itself.

   Concept and Intuition

   The whole point of the integrating factor $\mu(x)=e^{\int P\,dx}$ is that $\dfrac{d\mu}{dx}=P(x)\mu(x)$ (by the fundamental theorem of calculus applied to the exponent). This special property is exactly what's needed to make $\mu\dfrac{dy}{dx}+\mu P y$ collapse into a single product-rule derivative.

   Step-by-Step Solution

   1. Start with $\dfrac{dy}{dx}+P(x)y=Q(x)$ and multiply both sides by $\mu(x)=e^{\int P\,dx}$:

   $\mu\frac{dy}{dx}+\mu P(x)y=\mu Q(x).$

   2. Compute $\dfrac{d\mu}{dx}$: since $\mu=e^{\int P\,dx}$, by the chain rule $\dfrac{d\mu}{dx}=P(x)\cdot e^{\int P\,dx}=P(x)\mu$. …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The general solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1}y})\frac{dy}{dx} = 0$ is (A) $xe^{\tan^{-1}y} = \tan^{-1}y + c$ (B) $x^2e^{2\tan^{-1}y} = e^{\tan^{-1}y} + c$ (C) $(x-2) = ce^{-\tan^{-1}y}$ (D) $2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + c$

   ›Reveal solutionSolution

   This is a first-order linear ODE that becomes linear in $x$ (not $y$) once you recognise the $\tan^{-1}y$ integrating factor. The answer is (D).

   Concept and Intuition

   When the equation is not linear in $y$ but treating $x$ as the dependent variable (function of $y$) makes it linear, we should switch roles: write it as $\dfrac{dx}{dy}+P(y)x=Q(y)$ and use the integrating factor $e^{\int P\,dy}$.

   Step-by-Step Solution

   1. Given: $(1+y^2)+(x-e^{\tan^{-1}y})\dfrac{dy}{dx}=0$.
   2. Rearranging: $(x-e^{\tan^{-1}y})\dfrac{dy}{dx}=-(1+y^2)$, so $\dfrac{dx}{dy}=\dfrac{x-e^{\tan^{-1}y}}{-(1+y^2)}=\dfrac{e^{\tan^{-1}y}-x}{1+y^2}$.
   3. This gives the linear form: $\dfrac{dx}{dy}+\dfrac{1}{1+y^2}x=\dfrac{e^{\tan^{-1}y}}{1+y^2}$.
   4. Integrating factor: $I=e^{\int \frac{dy}{1+y^2}}=e^{\tan^{-1}y}$.
   5. Multiply through: $\dfrac{d}{dy}\left(x\,e^{\tan^{-1}y}\right)=e^{\tan^{-1}y}\cdot\dfrac{e^{\tan^{-1}y}}{1+y^2}=\dfrac{e^{2\tan^{-1}y}}{1+y^2}$.
   6. Let $t=\tan^{-1}y$, so $dt=\dfrac{dy}{1+y^2}$. Then $\int e^{2t}\,dt=\tfrac12 e^{2t}$. …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The integrating factor of the linear differential equation $\dfrac{dy}{dx} = \dfrac{1}{4x+3y}$ is (A) $e^{-4x}$ (B) $e^{4x}$ (C) $e^{3y}$ (D) $e^{-3y}$

   ›Reveal solutionSolution

   The ODE is not linear in $y$, but writing $x$ as the dependent variable turns it into $\dfrac{dx}{dy}-4x=3y$, whose integrating factor is $e^{-4y}$ — the option carrying the exponent $-4$.

   Concept and Intuition

   A first-order linear ODE must look like $\dfrac{dy}{dx}+P(x)y=Q(x)$. Here $\dfrac{dy}{dx}=\dfrac{1}{4x+3y}$ has the unknown buried in a denominator, so it is not linear in $y$. The rescue is a change of viewpoint: nothing stops us from regarding $x$ as the function and $y$ as the independent variable. Since $\dfrac{dx}{dy}=1\big/\dfrac{dy}{dx}$, the reciprocal instantly clears the denominator and the equation becomes linear in $x$. The integrating factor is then $e^{\int P\,dy}$, with $P$ the coefficient of $x$.

   Step-by-Step Solution

   1. Invert: $\dfrac{dx}{dy}=4x+3y$.
   2. Put it in standard linear form (dependent variable $x$):

   $\frac{dx}{dy}-4x=3y,\qquad P(y)=-4,\ \ Q(y)=3y .$

   3. Integrating factor: $\text{I.F.}=e^{\int P\,dy}=e^{\int(-4)\,dy}=e^{-4y}.$ …

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10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The general solution of the differential equation $(1 + \tan y)(dx - dy) + 2x\,dy = 0$ is (A) $e^x(y\cos x + \sin x) + \sin x = c$ (B) $e^x(y\cos x + y\sin x - \sin x) + \cos x = 0$ (C) $e^y(x\cos y + x\sin y - \sin y) = c$ (D) $e^y(x\cos y + x\sin y + \sin y) = c$

    ›Reveal solutionSolution

    This is a first-order linear ODE in $x$ as a function of $y$; finding the right integrating factor is the crux. Answer: option (C).

    Concept and Intuition

    Grouping the $dx$ terms and $dy$ terms shows this is linear in $x$ (treating $y$ as the independent variable), of the form $\frac{dx}{dy}+P(y)x=Q(y)$. The integrating factor $e^{\int P\,dy}$ simplifies neatly once we split $2\cos y/(\cos y+\sin y)$ using the identity for a sum/difference of sine and cosine.

    Step-by-Step Solution

    1. Expand: $(1+\tan y)dx - (1+\tan y)dy + 2x\,dy = 0 \Rightarrow (1+\tan y)dx + \left[2x-(1+\tan y)\right]dy=0$.
    2. Divide by $(1+\tan y)\,dy$: $\dfrac{dx}{dy} + \dfrac{2x}{1+\tan y} = 1$.
    3. Write $\dfrac{2}{1+\tan y} = \dfrac{2\cos y}{\cos y+\sin y}$. Using $2\cos y=(\cos y+\sin y)+(\cos y-\sin y)$: $\frac{2\cos y}{\cos y+\sin y}=1+\frac{\cos y-\sin y}{\cos y+\sin y}.$
    4. Integrating factor: $\mu(y)=\exp\!\left[\int\left(1+\frac{\cos y-\sin y}{\cos y+\sin y}\right)dy\right] = \exp\left[y+\log|\cos y+\sin y|\right] = e^y(\cos y+\sin y)$. …

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11. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.The solution of the differential equation $e^xy\,dx + e^x\,dy + x\,dx = 0$ is (A) $e^x+yx^2=c$ (B) $2ye^x+x^2=c$ (C) $ye^x+x^2e^y=c$ (D) $e^x+xe^y=c$

    ›Reveal solutionSolution

    The first two terms of the equation are exactly $d(ye^x)$; separating out the remaining $x\,dx$ term and integrating directly gives the solution.

    Concept and Intuition

    Many differential equations that look complicated are secretly "exact" — the left side is the total differential of some simple combination of $x$ and $y$. Spotting the pattern $u\,dv+v\,du=d(uv)$ (here with $u=y,\ v=e^x$) turns an equation that looks like it needs an integrating factor into a one-line integration.

    Step-by-Step Solution

    1. Given: $e^xy\,dx+e^x\,dy+x\,dx=0$.
    2. Recall $d(ye^x)=y\,d(e^x)+e^x\,dy=ye^x\,dx+e^x\,dy$ — exactly the first two terms of the given equation.
    3. So the equation becomes $d(ye^x)+x\,dx=0$, i.e. $d(ye^x)=-x\,dx$.
    4. Integrate both sides: $ye^x=-\dfrac{x^2}{2}+C$.
    5. Multiply through by 2: $2ye^x=-x^2+2C$, i.e. $2ye^x+x^2=c$ (writing $c=2C$).

    Common Mistakes …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.The general solution of the differential equation $x\log x\,dy = (x\log x - y)\,dx$ is (A) $(x - y)\log x + x = c$ (B) $x - y = \dfrac{x}{\log x} + c$ (C) $y - x = \dfrac{x}{\log x} + c$ (D) $(y - x)\log x + x = c$

    ›Reveal solutionSolution

    This tests recognizing a first-order LINEAR differential equation in disguise and applying the standard integrating-factor method.

    Concept and Intuition

    After dividing through by $x\log x$, the equation takes the standard linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$ with $P(x)=\dfrac{1}{x\log x}$. The integrating factor $e^{\int P\,dx}$ makes the left side an exact derivative $\dfrac{d}{dx}(y\cdot \text{IF})$, so the whole equation integrates directly.

    Step-by-Step Solution

    1. Divide both sides by $x\log x$: $\dfrac{dy}{dx}=1-\dfrac{y}{x\log x}\ \Rightarrow\ \dfrac{dy}{dx}+\dfrac{y}{x\log x}=1$.
    2. Integrating factor: $\mathrm{IF}=\exp\!\left(\int \dfrac{dx}{x\log x}\right)=\exp(\log(\log x))=\log x$ (since $\int\frac{dx}{x\log x}=\log(\log x)+C$).
    3. Multiply the ODE by $\log x$: $\dfrac{d}{dx}\big(y\log x\big)=\log x$. …

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